Constants Part A When an air capacitor with a capacitance of 330 nF (1 nF 109 F)

Question

Constants Part A When an air capacitor with a capacitance of 330 nF (1 nF 109 F) is connected to a power supply, the energy stored in the capacitor is 1.75x10-5J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.72x 10-5 J What is the potential difference between the capacitor plates? V-10.3 V Previous Answers Correct Part B What is the dielectric constant of the slab? K-1.55 SubmitP Incorrect; Try Again; 8 attempts remaining

Explanation / Answer

(A) U = C V^2 /2

(1.75 x 10^-5) = (330 x 10^-9) V^2 / 2

V = 10.3 Volt

(B) C' = k C

As it is connected to battery, PD will remain same

V= 10.3 Volt

U = C' V^2 / 2

((1.75 x 10^-5) + (2.72 x 10^-5)) = (k x 330 x 10^-9) (10.3^2) /2

k = 2.55

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